Construction of the Outer Automorphism of Via a Complex

Math.Comput.Sci. (2018) 12:453–458 https://doi.org/10.1007/s11786-018-0382-0 Mathematics in Computer Science

Construction of the Outer Automorphism of S6 via a Complex Hadamard Matrix

Neil I. Gillespie · Padraig Ó Catháin · Cheryl E. Praeger

Received: 15 December 2017 / Revised: 20 March 2018 / Accepted: 22 March 2018 / Published online: 26 September 2018 © The Author(s) 2018

Abstract We give a new construction of the outer automorphism of the symmetric group on six points. Our construction features a complex Hadamard matrix of order six containing third roots of unity and the algebra of split quaternions over the real numbers.

Keywords Complex hadamard matrix · Outer automorphism · Symmetric group

Mathematics Subject Classiﬁcation Primary 20B25; Secondary 05B20 · 20B30

1 Introduction

Sylvester showed that the ﬁfteen two-subsets of a six element set can be formed into 5 parallel classes in six different ways and that the action of S6 on these synthematic totals is essentially different from its natural action on six points [13]. To our knowledge this was the ﬁrst construction for the outer automorphism of S6. Miller attributes the result that for n = 6, Sn has no outer automorphisms to Hölder, and Sylvester’s construction of the outer automorphism of S6 to Burnside [11]. He also gives a by-hand construction of the outer automorphism. The papers of Janusz and Rotman, and of Ward provide easily readable accounts which are similar to Sylvester’s [10,14]. Cameron and van Lint devoted an entire chapter (their sixth!) to the outer automorphism of S6 [2]. They build on Sylvester’s construction to construct the 5-(12, 6, 1) Witt design, the projective plane of order 4, and the Hoffman–Singleton graph. Via consideration of the cube in R3, Fournelle gives a heuristic for the existence of an outer automorphism of S6, and constructs it with the aid of a computer [7]. Howard, Millson, Snowden and Vakil give two constructions of

N. I. Gillespie (B) Heilbronn Institute for Mathematical Research, University of Bristol, Tyndall Ave, Bristol BS8 1TH, UK e-mail: [email protected] P. Ó Catháin Department of Mathematical Sciences, Worcester Polytechnic Institute, 100 Institute Road, Worcester, MA 01609, USA e-mail: [email protected] C. E. Praeger School of Mathematics and Statistics, University of Western Australia, 35 Stirling Highway, Crawley, WA 6009, Australia e-mail: [email protected] 454 N. I. Gillespie et al. the outer automorphism of S6, and use this to describe the invariant theory of six points in certain projective spaces [9]. In this note we give a construction which we believe has not previously been described, using a complex Hadamard matrix of order 6 and a representation of the triple cover of A6 over the complex numbers. This note is inspired by a construction of Marshall Hall Jr [8] for the outer automorphism of M12 via a real Hadamard matrix of order 12, and by Moorhouse’s classiﬁcation of the complex Hadamard matrices with doubly transitive automorphism groups [12]. It was in the latter paper that we ﬁrst became aware of the complex Hadamard matrix of order 6 discussed in this article, where it is described as corresponding to the distance transitive triple cover of the complete bipartite graph K6,6.

2 Hadamard Matrices

Let ω be a primitive complex third root of unity. Then the matrix H6 is complex Hadamard. ⎛ ⎞ 111111 ⎜ ⎟ ⎜ 11 ω ω ωω⎟ ⎜ ⎟ ⎜ 1 ω 1 ω ω ω ⎟ H6 = ⎜ ⎟ ⎜ 1 ωω1 ω ω ⎟ ⎝ 1 ω ωω1 ω ⎠ 1 ω ω ωω1

† = † This means that H6 satisfies the identity H6 H6 6I6, where for an invertible complex matrix A, A is the complex conjugate transpose of A. Equivalently, H6 reaches equality in Hadamard’s determinant bound. We refer the reader to [6] for a comprehensive discussion of Hadamard matrices and their generalisations. An automorphism of a complex Hadamard matrix is a pair of monomial matrices (P, Q) such that P−1 HQ = H. The set of all automorphisms of H forms a group under composition. In this note we will work with the subgroup of automorphisms (P, Q) where all non-zero entries are third roots of unity, we denote this group Aut(H). Consider now the projection maps ρ (P, Q) → P and ρ (P, Q) → Q. Since √1 H is unitary, and for any automorphism 1 2 6 6 −1 (P, Q) of H the identity HQH = P holds, it follows that ρ1 and ρ2 are conjugate representations of Aut(H). Note further that ρi is a faithful representation, since Q = I forces P = I . Thus Aut(H) is isomorphic to a finite subgroup of monomial matrices of GLn(C). Furthermore, if Aut(H) contains a subgroup isomorphic to G, then the projections ρ1 and ρ2 onto the first and second components of Aut(H) give two conjugate representations of G by monomial matrices. Every monomial matrix has a unique factorisation P = DK where D is diagonal and K is a permutation matrix. The projection π : P → K is a homomorphism for any group of monomial matrices. In general, the representation ρ π ρ π Aut(H) 1 is not linearly equivalent to the representation Aut(H) 2 . As mentioned above, this phenomenon was first observed by Hall, who showed that the automorphism group of a Hadamard matrix of order 12 is isomorphic to 2 · M12, and that ρ1π and ρ2π realise the two inequivalent actions of M12 on 12 points [8]. This interpretation of the outer automorphism of M12 was also used by Elkies, Conway and Martin in their analysis of the Mathieu groupoid M13 [4]. Throughout this note we use the following shorthand for monomial matrices: we list the elements of the diagonal matrix D, and give the cycle notation for K as a permutation of the columns of the identity matrix (i.e. a right action). Consider the following pairs of monomial matrices.

τ1 := ([1, 1, 1, 1, 1, 1](2, 3, 4, 5, 6), [1, 1, 1, 1, 1, 1](2, 3, 4, 5, 6))

τ2 := ([1, 1,ω,ω, ω, ω](1, 2), [1, 1, ω,ω,ω,ω] (1, 2)(3, 6)(4, 5)) .

We deﬁne ∗ to be the entry-wise complex conjugation map, and consider the group X =τ1,τ2, ∗. Construction of the outer automorphism... 455

10 Proposition 1 The group X is of the form 3 · S6 · 2. τ ∗ = τ τ ∗ = τ −1 =τ ,τ = ∗ Proof Since 1 1 and 2 2 , we have that X0 1 2 is normal in X. Hence X X0 , with X0 of index 2 in X. The commutator [τ2, ∗] = ([1, 1,ω,ω, ω, ω], [1, 1, ω,ω,ω,ω]) consists of diagonal matrices; furthermore τ := [τ , ∗]−1τ = (( , ), ( , )( , )( , )), 2 2 2 1 2 1 2 3 6 4 5 a pair of permutation matrices. Recall that s, t | s6 = t2 = (st)5 =[t, s2]2 =[t, s3]2 = 1 is a presentation for S = τ 6 (see [1], for example). A computation with t 2 and = τ τ = (( , , , , , ), ( , , )( , )) s 1 2 1 2 3 4 5 6 1 2 6 3 5 , =τ ,τ shows that all the relations in this presentation hold for these elements s t, and hence Y 1 2 is isomorphic ρ π ∼ to a quotient of S6. On the other hand, Y 1 is easily seen to be isomorphic to S6, so we conclude that Y = S6. Now let N be the subgroup of X consisting of all elements for which each component is a diagonal matrix. Since ρ ρ ρ τ i and τ i have determinants in {±1}, every element of the projection X i also has determinant ±1. However all 1 2 0 ρ ρi i the elements of N have third roots of unity along the diagonal, and so must have determinant 1. As a result, X0 ∼ 5 is isomorphic to a subgroup of M S6 where M = 3 is the group of unimodular diagonal matrices with entries ρ π from ω, and S6 acts as Y i . The only non-trivial S6-submodule of M is the constant module of order 3. τ i Deﬁne ni+1 := [τ2, ∗] 1 for each i ≥ 1. (We shift subscripts because the action of τ1 on [τ2, ∗] gives elements of N which have the non-initial rows of H6 as the diagonal of the ﬁrst component.) Since [τ2, ∗] ∈ X0,wehave ni ∈ X0 for 2 ≤ i ≤ 6. Observe that 2 2 = ([ , , , ,ω,ω], , , , , ω, ω ) n3n4n5 1 1 1 1 [1 1 1 1 ] τ ( 2 2) 2 = ([ , , , ,ω,ω], , ,ω,ω, , ) . n3n4n5 1 1 1 1 [1 1 1 1] ρ ρ ρ So neither of the projections N 1 , N 2 are onto the constant module, and the kernel of N 1 is neither trivial nor the constant module. It follows that N =∼ M × M. Finally, we observe that monomial matrices normalise diagonal matrices, and that X0 acts as a group of monomial matrices in each component. It follows that N X0, and that Y is a complement of N in X0. Since ∗ acts on N by inversion, N X.

(P,Q) −1 The group X has a natural action on 6 × 6 matrices over C where (P, Q) ∈ X0 acts as H = P HQ, and ◦ ∗ acts by complex conjugation. We compute the stabiliser of H6 under this action. We denote this group Aut (H6) to emphasise that this is a group of semi-linear transformations in its action on the normal subgroup N. We require the subgroups X0, Y and N defined in Proposition 1 in the proof of the following. ◦ ◦ Proposition 2 The group Aut (H6) is isomorphic to the nonsplit extension 3·S6, and Aut (H6) contains a C-linear subgroup isomorphic to 3 · A6. τ τ 1 = 2 ∗ τ Proof It is easily verified by hand that H6 H6 while H6 is the complex conjugate H6 . Therefore both 1 and ◦ the product τ2∗ fix H6. We claim that Aut (H6) =τ1,τ2∗. ◦ First, we show that the intersection Aut (H6) ∩ N has order 3. To prove this, suppose that (D, E) ∈ N, and that −1 D H6 E = H6, or equivalently DH6 = H6 E. Since the first column of H6 is constant, D must be a scalar matrix. i i So D commutes with H6, and we have DH6 = H6 D = H6 E. Hence D = E,so(D, E) = (ω I,ω I ) for some i. Since these elements do leave H6 invariant, the claim is proved. ( , ) ∗ = We next claim that there is no element D E of N such that DH6 H6 E; suppose to the contrary that such a ( , ) ∗ = −1 D E exists. Precisely the same argument as before shows that D must be scalar. This implies that H6 H6 ED , ∗ but this equation has no solution in diagonal matrices: since the first row of H6 is equal to the first row of H6,we −1 = = ∗ would require ED I6, from which we derive H6 H6 , a contradiction. Consider the subgroup K := τ1,τ2∗, N of X. Since X =K, ∗ and ∗ ∈/ K ,wehave|X : K |=2 and = ∪ ( ∗) ∗ X K K . It follows, moreover, from the previous arguments that no element of K sends H6–H6 , and hence 456 N. I. Gillespie et al.

◦ no element of the right coset K ∗ can ﬁx H6. Therefore, Aut (H6) ⊆ K , and from the ﬁrst paragraph of the proof we ◦ ◦ ◦ also have Aut (H6)N = K . The quotient Aut (H6)/(Aut (H6) ∩ N) is isomorphic to K/N, an index 2 subgroup ∼ of X/N = S6 · 2. In particular K/N contains A6 as a normal subgroup of index 2. Since the element Nτ2∗ does ∼ not lie in A6 and does not centralise A6 it follows that K/N = S6. ◦ We have shown that Aut (H6) has a normal subgroup of order 3 with quotient isomorphic to S6. The elements τ i (τ2∗) 1 for 0 ≤ i ≤ 4 project onto a set of Coxeter generators for S6. With these generators, it is straightforward to ◦ construct a Sylow 3-subgroup of Aut (H6). One such subgroup is generated by x := ([ω, 1,ω,ω,1, ω](1, 2, 3), [ω,1,ω,1, ω, ω](1, 4, 6)(2, 3, 5)) y := ([ω,ω, 1, 1, ω, ω](4, 5, 6), [ω,ω,ω,ω,ω,ω](1, 4, 6)(2, 5, 3)) . A computation shows that [x, y] = ([ω,ω,ω,ω,ω,ω], [ω,ω,ω,ω,ω,ω]). This shows that the commutator subgroup contains the normal subgroup of order 3, hence the extension is non-split. Elements of Aut◦(H) which map onto odd permutations act on [x, y] by inversion. So the centraliser of this normal subgroup is of index 2 in ◦ Aut (H): this is necessarily a non-split central extension 3 · A6. A perfect group S has a largest non-split central extension Sˆ which is unique up to isomorphism. The center of Sˆ is the Schur multiplier of S, and every non-split central extension of S is a quotient of Sˆ. The number of generators of the Schur multiplier is bounded by g − r where g is the number of generators in a presentation of S and r is the number of relations. We refer the reader to Wiegold’s survey on the Schur multiplier for proofs of all these results [15]. Since A6 is shown in [3] to have the presentation a, b | a4, b5, abab−1abab−1a−1b−1 , it follows that the Schur multiplier of A6 is cyclic. Hence the non-split extension 3.A6 is unique up to isomorphism. ◦ ◦ Now, since Aut (H) splits over 3· A6, we have that 3· A6 < Aut (H)

◦ In fact, 3.A6 is the largest subgroup of Aut (H6) admitting a faithful 6-dimensional representation over C.So ◦ this is Aut(H6). A useful way to understand the actions of X and of Aut (H6) is via a permutation action on 18 ρ ρ = τ 1 = τ 1 × points, which we now describe. Let P1 1 and P2 2 , and deﬁne the following 18 6 matrices: ⎛ ⎞ ⎛ ⎞ H H ∗ ⎝ ⎠ ⎝ ∗ ⎠ M1 = ωH and M2 = ωH . ω2 H ω2 H ∗

For 1 ≤ i ≤ 18, let Rowi (M j ) denote the ith row of M j (where j = 1, 2). Let P1 act on the rows of M1, and similarly the rows of M2, as follows: ⎛ ⎞ P1 H ⎝ ⎠ P1 · M1 = ωP1 H 2 ω P1 H

By letting P2 act on the rows of M1 and M2 in a similar manner, we ﬁnd that P1 and P2 act in the same way on the rows of M1 and the rows of M2, and hence act on the set (18) := {{Rowi (M1), Rowi (M2)}|i = 1,...,18}. Also, letting ∗ act as complex conjugation on M1 and M2, we see that ∗ also induces a permutation of (18). Thus Construction of the outer automorphism... 457

τ1, τ2 and ∗ all induce permutations of (18) and, identifying {Rowi (M1), Rowi (M2)} with i, for each i, we get a permutation representation of X on 18 points with the following generating permutations:

τ1 = (2, 3, 4, 5, 6)(8, 9, 10, 11, 12)(14, 15, 16, 17, 18),

τ2 = (1, 2)(3, 15, 9)(4, 10, 16)(5, 11, 17)(6, 18, 12)(7, 8)(13, 14), ∗=(7, 13)(8, 14)(9, 15)(10, 16)(11, 17)(12, 18). The kernel of X in this action is the subgroup of N of order 35 consisting of pairs with trivial first component. The ◦ restriction to Aut (H6) is faithful, however. One could construct a faithful action of X by taking the permutation action induced by its action on the rows of H6 together with the induced action on columns. × Remark 3 The matrix H6 and the group 3.A6 can be realised over any field k for which k has a subgroup of order 3. In the case that k is the finite field of order 4, the rows of H6 span the Hexacode, introduced by Conway as part of a construction for the group M12. It is discussed in detail in Sect. 11.2 of [5]. In particular, this code is the extended quadratic residue code with parameters (6, 3, 4). Uniqueness can easily be verified by hand: observe that the punctured code is the Hamming (5, 3, 3) code, which is unique, and that any pair of one-bit extensions which increase the minimum distance are isomorphic. The 6-dimensional C-representation of 3 · A6 has been previously described in the literature, normally via its action on a set of vectors in C6 derived from the hexacode. In particular, Wilson gives the action of 3 · A6 on certain vectors of weight 4 in Sect. 2.7.4 of [16].

3 The Outer Automorphism of S6

Finally we construct the outer automorphism of S6 over the split-quaternions. Recall that the split-quaternions are a 4-dimensional R-algebra with basis [1, i,β,βi] where [1, i] generates the usual algebra of complex numbers and β2 = 1, iβ =−i. We denote the split quaternions by B. They admit an R-linear representation generated by 0 −1 01 i → ,β→ . 10 10 ◦ Observe that Aut (H6) admits a B-linear representation if and only if ∗ does, and that the latter is realised by (β I6,βI6). τ ∗ C B 2 = Since H6 is invertible over , it is invertible over . Now, rearranging the matrix equation H6 H6, and using the same notation as before for monomial matrices, we obtain that β,β,βω, βω, βω, βω ( , )( , )( , ) −1 = β,β,βω,βω, βω, βω ( , ) . H6 [[ ] 1 2 3 6 4 5 ] H6 [[ ] 1 2 ] Note that (βω)2 = (βω)2 = 1 so that the matrix on the right hand side of the above equation is an involution. As was the case over the complex numbers, H6 intertwines the projections ρ1 and ρ2. We observe that for any ◦ ρ ρ − ρ π ρ π ∈ ( ) 1 = 2 1 τ 1 τ 2 g Aut H , we have that g H6g H6 . But, as illustrated above, 2 is a 2-cycle, while the projection 2 is a product of 3 disjoint 2-cycles. We conclude that the representations ρ1π and ρ2π of S6 cannot be conjugate. Thus whereas the permutation representations of S6 on 6 points are not equivalent, and the monomial representations of 3 · A6 are not equivalent, we have constructed two explicit B-linear representations of 3 · S6 which are equivalent under conjugation by H6. Moreover, although the representation is not deﬁned over C, the intertwiner H6 is.

Theorem 4 There exists an irreducible 6-dimensional monomial representation of 3·S6 over the split-quaternions. Two conjugate representations of 3·S6 intertwined by the complex Hadamard matrix H6 give an explicit construction for the outer automorphism of S6.

Acknowledgements Work on this paper was begun while the second author was visiting the Centre for the Mathematics of Symmetry and Computation at the University of Western Australia in March 2012. The hospitality of the CMSC is gratefully acknowledged, and in particular support from the ARC Federation Fellowship Grant FF0776186 of the third author, which also supported the ﬁrst author. The second author acknowledges the support of the Australian Research Council via grant DP120103067, and Monash University where much of this work was completed. This research was partially supported by the Academy of Finland (Grants #276031, #282938, #283262 and #283437). The support from the European Science Foundation under the COST Action IC1104 is also gratefully acknowledged. 458 N. I. Gillespie et al.

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